Does this series converge, and if so to what value?: $\sum_{n=0}^\infty \left\{\frac{1}{(n+1)^2} \right\}\ln(2n+1)$ I've arrived at this series from a given sequence of terms, but now I'm at a loss as to how to proceed... How does one know which convergence test to use? This isn't a geometric series, so I don't have a handy formula to plug numbers into. And taking the limit as $x\ \to \infty$ gives me an indeterminate form of $0 \cdot \infty$. 
$$\sum_{n=0}^\infty \left\{\frac{1}{(n+1)^2} \right\}\ln(2n+1)$$
 A: To prove convergence:
For any $\varepsilon > 0$, we have $\ln (2n+1) < n^{\epsilon}$ for all $n > n_{\varepsilon}$ for a certain $n_{\varepsilon}$ depending on $\varepsilon$ (hence the subscript)
So, let $\varepsilon = \frac{1}{2}$, then we can compare this series to $\displaystyle \sum_{n=0}^{\infty} \frac{1}{(n+1)^{1.5}}$, which converges by the $p$-series test since $p = 1.5$ is greater than $1$.
Note that finding the exact value of a sum is often much more difficult than proving whether or not it converges. I calculated up to the $100000$th partial sum and got $S \approx 1.2713728 \dots$
A: Use equivalents: $\;\ln(2n+1)\sim_{\infty} \ln n$, $\;(n+1)^2\sim_{\infty}n^2$, so $\,\dfrac{\ln(2n+1)}{(n+1)^2}\sim_{\infty}\dfrac{\ln n}{n^2}$, which is a convergent Bertrand's series.
Note: A Bertrand's series is a series  $\;\displaystyle\sum\limits_n\dfrac1{n^\alpha(\ln n)^\beta}$. It converges if and only if $\alpha>1$ or $\,(\alpha=1$ and $\beta>1)$. It can be proved using the integral test.
A: To show that the series converges, notice that
$$
\begin{align}
\sum_{n=0}^\infty\frac{\log(2n+1)}{(n+1)^2}
&\le\sum_{n=0}^\infty\frac{\log(2n+2)}{(n+1)^2}\\
&=\sum_{n=1}^\infty\frac{\log(2n)}{n^2}\\
&=\log(2)\color{#00A000}{\sum_{n=1}^\infty\frac1{n^2}}+\color{#C00000}{\sum_{n=1}^\infty\frac{\log(n)}{n^2}}\\
\end{align}
$$
We can show that both of these sums converges by the integral test. Integrating by parts yields
$$
\begin{align}
\color{#C00000}{\int_1^\infty\frac{\log(x)}{x^2}\,\mathrm{d}x}
&=-\int_1^\infty\log(x)\,\mathrm{d}\frac1x\\
&=\int_1^\infty\frac1x\,\mathrm{d}\log(x)\\
&=\color{#00A000}{\int_1^\infty\frac1{x^2}\,\mathrm{d}x}\\[6pt]
&=1
\end{align}
$$
Using the Euler-Maclaurin Sum Formula to $O\!\left(\log(n)n^{-21}\right)$ and summing to $n=100$, we get
$$
\sum_{n=0}^\infty\frac{\log(2n+1)}{(n+1)^2}\doteq1.271388320211200139286593230121
$$
A: Hint:
The very similar series
$$\sum_{n=0}^\infty \frac{\ln(2n+\color{red}2)}{(n+1)^2}=\sum_{n=0}^\infty \frac{\ln(n+1)+\ln(2)}{(n+1)^2}=-\zeta'(2)+\ln(2)\zeta(2).$$
A correction can be brought with 
$$\ln\left(\frac{2n+1}{2n+2}\right)=\ln\left(1-\frac1{2n+2}\right)=-\sum_{k=1}^\infty\frac1{k(2n+2)^k},$$
leading to
$$-\sum_{k=1}^\infty\frac{\zeta(k+2)}{2^kk}.$$
